### Recent Posts

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##### Chapter 3 / Re: Chapter 3.1 Theorem 4
« Last post by Victor Ivrii on February 27, 2022, 07:52:52 PM »
Indeed, to be corrected. Thanks
52
##### Chapter 3 / Chapter 3.1 Theorem 4
« Last post by Zicheng Ding on February 26, 2022, 07:43:05 PM »
In the online text book, chapter 3.1 theorem 4, we have the formula for inhomogeneous heat equations. For the second integral in the formula, since it represents the homogeneous part of the heat equation and the heat equation is linear, should the range be from $-\infty$ to $\infty$ instead of $0$ to $\infty$? The equation we are solving have the range of $-\infty < x < \infty$ and $t > 0$ as usual.
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##### Test 1 / 2020 Test 1 Q2
« Last post by kevin on February 23, 2022, 08:45:59 PM »
how did we derive 2 from 1 and 4 from 3
54
##### Chapter 2 / Textbook Chapter 2.5 Example 2
« Last post by Yifei Hu on February 21, 2022, 11:29:41 AM »
Hi Professor,

I think one coefficient is wrong in the solution.

In this problem the solution to the inhomogeneous equation with homogeneous initial condition should start with coefficient $\frac{1}{2c} = \frac{1}{6}$. After cancelling the 36 in the $f(x,t)$, we should arrive at $6\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$ instead of $3\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$.

Can you help me confirm if I missed something or the textbook has a typo?
55
##### Chapter 2 / Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Last post by Victor Ivrii on February 21, 2022, 04:34:15 AM »
Quote
But how does this qualify us to replace the indefinite integral with the definite one?

Did you take Calculus I? Then you must know that if the preimitive (indefinite integral) is a set of definite integrals which differ by an arbitrary constant.
56
##### Chapter 2 / Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Last post by Yifei Hu on February 20, 2022, 08:12:28 PM »
I understand that I can show $\phi'(\xi)=0$ by:

1) when t = 0, $\xi = x+ct = x = x-ct = \eta$
2) $u_\xi = u_t \frac{dt}{d\xi} + u_x \frac{dx}{d\xi}$ by chain rule.
3)By initial condition: $u_t|_{t=0} = u_x|_{t=0}$ = 0, we must have $u_\xi=0$
4) $u_\xi = \phi'(x)$ hence $\phi'(\xi)=0$

But how does this qualify us to replace the indefinite integral with the definite one?
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##### Chapter 2 / Re: Text Book 2.4 Example 2.1
« Last post by Victor Ivrii on February 20, 2022, 10:04:48 AM »
We integrate from $t=0$ because for $t=0$ initial conditions are done. $-1<t$ is a domain where $f(x,t)$ is defined
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##### Chapter 2 / Re: Derivation of D' Alembert formula under Characteristic Coordinate
« Last post by Victor Ivrii on February 20, 2022, 10:01:28 AM »
Reproduce formula correctly (there are several errors) and think about explanation why $\phi'(\xi)=0$.
59
##### Chapter 2 / Text Book 2.4 Example 2.1
« Last post by Yifei Hu on February 19, 2022, 09:16:55 PM »
In this problem, $-1 < t < -\infty$, when applying D' Alembert's formula, why we integrate from 0 to t instead of from -1 to t?
In text book 2.4, we had one line in the derivation: $\tilde{u_{\xi}}=-\frac{1}{4c^2} \int^\xi f(\xi,\eta')d\eta'=-\frac{1}{4c^2} \int_\xi^\eta\tilde f(\xi,\eta')d\eta' + \phi'(\xi)$.
Why can we replace the definite integral with the indefinite integral? Why we choose $\xi$ as lower limit and $\eta$ as upper limit?